Ph-CH(OH)-CH_3 xrightarrow{PCC} (A) xrightarr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes secondary alcohols to ketones. Thus,$(A)$ is acetophenone $(Ph-C(=O)-CH_

Vedclass · Accepted Answer

$Ph-C(CH_3)=N-NH-C(=O)-NH_2$

Vedclass · Answer

(B) $PCC$ (Pyridinium chlorochromate) is a mild oxidizing agent that oxidizes secondary alcohols to ketones. Thus,$(A)$ is acetophenone $(Ph-C(=O)-CH_3)$. Acetophenone reacts with semicarbazide $(NH_2-NH-C(=O)-NH_2)$ to form acetophenone semicarbazone. In semicarbazide,the $-NH_2$ group attached to the carbonyl group is less nucleophilic due to resonance,so the terminal $-NH_2$ group of the hydrazine part reacts with the carbonyl group of the ketone. The product $(B)$ is $Ph-C(CH_3)=N-NH-C(=O)-NH_2$.

$Ph-CH(OH)-CH_3$ $\xrightarrow{PCC} (A)$ $\xrightarrow{NH_2-NH-C(=O)-NH_2} (B)$. Product $(B)$ is:

Similar Questions

Which of the following will fail to react with potassium dichromate and dilute sulphuric acid?

Addition of alcohols to aldehydes in the presence of anhydrous acids yields:

Which of the following compounds does not undergo the Cannizzaro reaction?

The reagent that can distinguish benzaldehyde from propionaldehyde is

Silvering of a mirror is done by using which of the following compounds?

Vedclass Test Series

Exam Paper Generator

Online Exam Module